Question

Let a, b be integers st (3,a) = (3, b) =1. Show that a^2 + b^2 is not the square of an integer.
Please, help

Answer 1

(3,a) = 1 means that the GCD of 3 and a is 1
So a does not equal 3 or any of its multiples
which means a = 3k+1 or a = 3k+2


Similar logic applies to b as well since (3,b) = 1 too.

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You'll need to break things down into 3 cases


Case 1: a = 3k+1, b = 3k+1
Case 2: a = 3k+1, b = 3k+2
Case 3: a = 3k+2, b = 3k+2


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Proving Case 1


a = 3k+1
b = 3k+1
k is any integer


a^2 + b^2 = (3k+1)^2 + (3k+1)^2
a^2 + b^2 = 2*(3k+1)^2


2*(3k+1)^2 is NOT in the form m^2, where m is an integer. Why not? The portion (3k+1)^2 is certainly a perfect square but the 2 out front is not a perfect square


Put another way


m^2 = 2*(3k+1)^2
m = sqrt(2*(3k+1)^2)
m = sqrt(2)*sqrt((3k+1)^2)
m = sqrt(2)*(3k+1)

So m is irrational and not an integer. This proves a^2 + b^2 is not a perfect square if a = 3k+1, b = 3k+1, where k is any integer

Answer 2

Alternate solution:
\[a=3n \pm 1 \]\[b=3m \pm 1\]
\[a^2 + b^2 \equiv c^2 \mod 3 \]\[(\pm 1)^2 + (\pm 1)^2 \equiv c^2 \mod 3\]\[2 \equiv c^2 \mod 3\]

But in mod 3, \(0^2=0\), \(1^2=1\) and \(2^2=1\), so contradiction.

Answer 3

I'm not thinking. I should have used different integers for a and b like @Empty did. My bad

Answer 4

Yeah, I just wanted to have some fun with this, my first instinct was to do the exact same thing you did, but I wanted to find a better way!

Answer 5

I got @jim_thompson5910 method
@Empty is it the same, but combine the cases, right? it doesn't derive from (a, 3) =1, right?

Answer 6

Yeah this is the same thing as what he did, just in a more convenient notation that allows us to do all the cases at one time because \(a=3n \pm 1\) is really the same as \(3n+1\) and \(3n+2\). Does that answer your question? I don't completely understand your question

Answer 7

Got it. Thanks a lot. It is not my problem. Since my second midterm is upcoming, I search on internet problems to practice. It doesn't have solution. Hence when I am not sure about the solution, I put it here to learn. http://math.scu.edu/~eschaefe/NTpp1.pdf
Can you take a look at problem 11? Please

Answer 8

To me, we solve 27x - my = 5
gcd (27, m) = d and d | 5 iff d =1 or =5, but 5 not | 27, hence d =1

Answer 9

I think you should remove the x from your equation and just solve \(27=5+my\)

Answer 10

but then, how to solve for m?

Answer 11

I have to go eat dinner, I'll be back later, but I would just look at the prime factorizations. There might be more solutions than just this, I need to think a little more about this but you should be able to see that m=2 and m=11 will work I think.

Answer 12

if (27,m) =1, and 27 = 3^3 , hence residue {1,2......26} without multiple of 3. right?

Answer 13

a = b (mod m) means m | (a-b)
27 = 5 (mod m) means that m | (27-5) ---> m | 22

so m*k = 22 meaning that m is a factor of 22 (k is an integer)

so if m = 2, m = 11, or m = 22, then 27 = 5 (mod m)


m = 2:
27 = 5 (mod m)
27 = 5 (mod 2)
1 = 1 (mod 2)


m = 11:
27 = 5 (mod m)
27 = 5 (mod 11)
5 = 5 (mod 11)


m = 22:
27 = 5 (mod m)
27 = 5 (mod 22)
5 = 5 (mod 22)


I think those are all the possible values of m but I'm not 100% sure

Answer 14

Wow!! it is so simple. Thank you Mr. Jim

Answer 15

no problem

Answer 16

Mr. Jim, how to do 13? Please